Show that for all $m\ge 2$ there is some invertible, non-diagonalizable matrix $A\in M_2(\mathbb{R})$ s.t $A^m$ is diagonalizable.

Question

Show that for all $m\ge 2$ there is some invertible, non-diagonalizable matrix $A\in M_2(\mathbb{R})$ s.t $A^m$ is diagonalizable.

Hey. This statement is false over $\mathbb{C}$, but over $\mathbb{R}$ we can consider the rotation matrix $$ \begin{pmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{pmatrix} $$ Note that for $\theta=\frac{\pi}{2}$ we obtain $A= \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$ which is an invertible, non-diagonalizable matrix, but $A^2 = -I \implies A^{2k}=I\ \lor A^{2k}=-I \ \forall k\in\mathbb{N} $ so for even powers I have found such a matrix (although I am not sure this is how I am supposed to prove this).

Now we notice that for $\theta=\frac{\pi}{3}$ we obtain $A= \begin{pmatrix} \frac{1}{2} & \frac{\sqrt3}{2}\\ -\frac{\sqrt3}{2} & \frac{1}{2} \end{pmatrix}$ which is a matrix as listed above that satisfies $A^3=-I$.

Note that when $\theta=\frac{\pi}{5}$ we get a matrix that satisfies the given conditions, with $A^5=-I$. So there's a pattern here, I kind of see the geometric interpretation of this, although I am not sure this is the proper way to prove this claim.

I would be happy to hear your thoughts :)

Answer 1

Well, for $m \ge 2$, if you consider the rotation $A$ of angle $\frac{2 \pi }{m}$, it is certainly inversible and not diagonalisable. However, $A^m=I$.

Answer 2

Hint

What about taking for $A$ the matrix of the rotation of angle $\frac{2\pi}{m}$?